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How to solve the equation $x + \sin x = A$?

I have tried Wolfram Alpha, with no success.

I would be satisfied with an answer in special functions.

The context is, I was trying to find an expression for the coordinates $(x,y)$ for a number given with modulus and argument in the second system in MO post #423657.

I also tried to solve Reduce[{(a^2 + b^2)/a == r, 1/4 (ArcTan[b/a] + (a b)/(a^2 + b^2) ) == arg}, {a, b}] in Mathematica, but it took 10 hours after which I manually aborted.

Calvin Khor
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Anixx
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    For a specific value of $A$, probably numerically. – ncmathsadist Jun 06 '22 at 18:10
  • Try this inverse function. Also what is the context of the problem and what have you tried? Your question may be closed for these reasons. – Тyma Gaidash Jun 06 '22 at 18:13
  • @TymaGaidash Hello! The context is, I was trying to find an expression for the coordinates $(x,y)$ for a number given with modulus and argument in the second system in this post: https://mathoverflow.net/questions/423657/lemniscate-numbers-and-others-what-would-be-the-properties – Anixx Jun 06 '22 at 18:18
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    @TymaGaidash in that system, $\int_0^{\phi } \cos ^2(t) , dt=\operatorname{arg} z$. So to find the coordinates we have to find the angle $\phi$ from known arg. – Anixx Jun 06 '22 at 18:20
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    $f(x)=x+\sin x$ is increasing since $f'(x)=1+\cos x \geq 0$. It follows that there is a unique solution of $x+\sin x=A$ and it lies in the inverval $[A-1,A+1]$. Once you restrict to such interval, you may apply a numerical method like the secant method or the secant-tangent method. – Jack D'Aurizio Jun 06 '22 at 18:21
  • @JackD'Aurizio definitely, a numerical method should work, but I want a closed form for a post here about the properties of this numerical system. If I post without a closed form, as I did, I get downvotes. – Anixx Jun 06 '22 at 18:23
  • I also tried to solve Reduce[{(a^2 + b^2)/a == r, 1/4 (ArcTan[b/a] + (a b)/(a^2 + b^2) ) == arg}, {a, b}] in Mathematica, but it took 10 hours after which I manually aborted. – Anixx Jun 06 '22 at 18:24
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    If you apply the Lagrange inversion theorem you can write the inverse function as a power series, with the caveat that its coefficients are increasingly harder to be computed. – Jack D'Aurizio Jun 06 '22 at 18:25
  • @TymaGaidash Wow! that function in your post is indeed on spot! – Anixx Jun 06 '22 at 18:26
  • @TymaGaidash okay, so if I am not mistaken, $x=\frac{\sqrt{1-I_{\frac{4 \arg (z)}{\pi }}^{-1}\left(\frac{1}{2},\frac{3}{2}\right){}^2} \sqrt{\left| 1-I_{\frac{4 \arg (z)}{\pi }}^{-1}\left(\frac{1}{2},\frac{3}{2}\right){}^2\right| }}{\arcsin \left(I_{\frac{4 \arg (z)}{\pi }}^{-1}\left(\frac{1}{2},\frac{3}{2}\right)\right)}$ and $y=\frac{I_{\frac{4 \arg (z)}{\pi }}^{-1}\left(\frac{1}{2},\frac{3}{2}\right) \sqrt{\left| 1-I_{\frac{4 \arg (z)}{\pi }}^{-1}\left(\frac{1}{2},\frac{3}{2}\right){}^2\right| }}{\arcsin\left(I_{\frac{4 \arg (z)}{\pi }}^{-1}\left(\frac{1}{2},\frac{3}{2}\right)\right)}$ – Anixx Jun 06 '22 at 18:40
  • @TymaGaidash https://www.wolframalpha.com/input?i=solve+x%2Bcos+x+sin+x%3D%3DA – Anixx Jun 06 '22 at 18:48
  • @TymaGaidash it is from your post. I will make a new post about this algebraic system with divisors of infinity! – Anixx Jun 06 '22 at 18:57
  • Let us continue this discussion in chat. – Тyma Gaidash Jun 06 '22 at 19:08

1 Answers1

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Here is the solution for:

$$\int_0^\phi \cos^2(t) dt=\frac14(2\phi+\sin(2\phi))=\arg(z)$$

Therefore:

$2\phi+\sin(2\phi)=4\arg(z)$

Using

Inverse of $\sin(x)+x$

the result is:

$$\boxed{\phi=\frac12\text{hav}^{-1}\left(\text I^{-1}_{\frac{4\arg(z)}\pi}\left(\frac12,\frac32\right)\right)}$$

if the argument is satisfied for

$0\le \frac{ab}{a^2+b^2}+\text{sgn}(b)\text{Abs}\left(\tan^{-1}(a,b)\right)\le \pi$ for $\arg(z)=\arg(a,b)= \frac14\left(\frac{ab}{a^2+b^2}+\text{sgn}(b)\text{Abs}\left(\tan^{-1}(a,b)\right)\right)$

then the formula

works

Тyma Gaidash
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